Solve x³ - x² - 21x + 45 = 0
The solutions of x³ - x² - 21x + 45 = 0 are x = -5, 3, 3 (repeated roots counted once each: -5, 3).
How to work it out
- 1
A cubic has up to 3 real roots. The rational root theorem says that any integer root of a polynomial with integer coefficients and leading coefficient 1 divides the constant term 45, so the candidates are ±1, ±3, ±5, ±9, ±15, ±45.
- 2
Test the candidates until one gives 0. x = 3 is a root. Divide by (x − 3) with synthetic division; the bottom row is the quotient and a 0 remainder.
- 3
What is left is the quadratic x² + 2x - 15 = 0. Solve it (factoring or the quadratic formula):
- 4
So the polynomial factors as
LaTeX source of the formulas on this page
\[ x^{3} - x^{2} - 21x + 45 = (x + 5)(x - 3)^{2} = 0 \]
\[ x^{3} - x^{2} - 21x + 45 = 0 \]
\[ \begin{array}{c|cccc} 3 & 1 & -1 & -21 & 45 \\ & 0 & 3 & 6 & -45 \\ \hline & 1 & 2 & -15 & 0 \end{array} \]
\[ x^{2} + 2x - 15 = 0 \Rightarrow x = 3,\ -5 \]
\[ x^{3} - x^{2} - 21x + 45 = (x + 5)(x - 3)^{2} \]Candidates tested (computed)
| Candidate x | p(x) | Root? |
|---|---|---|
| 1 | 24 | no |
| -1 | 64 | no |
| 3 | 0 | root |
Vieta's formulas check the answer (computed)
| Relation | From the roots | From the coefficients |
|---|---|---|
| Sum of the roots | 1 | 1 |
| Product of the roots | -45 | -45 |
Questions
What are the roots of x³ - x² - 21x + 45 = 0?
The solutions of x³ - x² - 21x + 45 = 0 are x = -5, 3, 3 (repeated roots counted once each: -5, 3).
How do you factor x³ - x² - 21x + 45?
(x + 5)(x − 3).
What is the sum of the roots?
1 (it equals −(-1) by Vieta's formulas).